Answer:
B
Explanation:
triple beam balance
sorry if i am wrong! can be brainliest?
Answer:
The density of beryllium is 1.85 g/cm³. Thus given metal is beryllium.
Explanation:
Given data:
Mass of metal = 37 g
Volume of metal = 20cm³
Which metal is this = ?
Solution:
We will solve this problem by using density formula. After finding density we will check from literature which metal has a density similar to our calculated density.
d = m/v
d = 37 g/ 20cm³
d = 1.85 g/cm³
The density of beryllium is 1.85 g/cm³. Thus given metal is beryllium.
Answer:
The correct answer is Option C (E1) and Option B (carbocation).
Explanation:
- Intramolecular immunity idols are considered as that of the formation mechanism with E1 responses or reactivity.
- Reactants with E1 were indeed obligations of both parties, meaning that an E1 reaction was conducted thru all the two stages known as ionization but rather deprotonation. Involves the absence of either an aromatic ring, a carbocation has been generated throughout the ionization solution.
Some other possibilities offered aren't relevant to the procedure outlined. So the above alternative is accurate.
Answer:
Where is the diagram?please put the pictu
Answer:
Explanation:
In general, an increase in pressure (decrease in volume) favors the net reaction that decreases the total number of moles of gases, and a decrease in pressure (increase in volume) favors the net reaction that increases the total number of moles of gases.
Δn= b - a
Δn= moles of gaseous products - moles of gaseous reactants
Therefore, <u>after the increase in volume</u>:
- If Δn= −1 ⇒ there are more moles of gaseous reactants than gaseous products. The equilibrium will be shifted towards the products, that is, from left to right, and K>Q.
- If Δn= 0 ⇒ there is the same amount of gaseous moles, both in products and reactants. The system is at equilibrium and K=Q.
- Δn= +1 ⇒ there are more moles of gaseous products than gaseous reactants. The equilibrium will be shifted towards the reactants, that is, from right to left, and K<Q.
